Question

Convex lens and concave lens are placed at a separation of 5cm each of f=10 cm . If an object is placed at a distance of 30 cm infront of convex lens . Find the position and nature of the final image formed

Answer 1

Given,

$Sepration=5cm$

$f=10cm$

$u=30cm$

So, For convex lens:

$u_1=-30cm$

$f_1=+10cm$

$v_1=?$

By using lens formula:

$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u_1}$

$\frac{1}{10}=\frac{1}{v_1}+\frac{1}{30}$

$\frac{1}{v_1}=\frac{1}{10}-\frac{1}{30}$

$v=15cm$

The image act virtual image for the concave lens.

For convex lens:

$u_2=+10cm$

$f_2=-10cm$

$v_2=?$

By using lens formula:

$\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}$

$-\frac{1}{10}=\frac{1}{v_2}-\frac{1}{10}$

$\frac{1}{v_2}=\frac{1}{10}-\frac{1}{10}$

$v=0$

Thus, the final image formed at infinity.